PAT A1082. Read Number in Chinese (25/21)

Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese way. Output "Fu" first if it is negative. For example, -123456789 is read as "Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu". Note: zero ("ling") must be handled correctly according to the Chinese tradition. For example, 100800 is "yi Shi Wan ling ba Bai".

Input Specification:

Each input file contains one test case, which gives an integer with no more than 9 digits.

Output Specification:

For each test case, print in a line the Chinese way of reading the number. The characters are separated by a space and there must be no extra space at the end of the line.

Sample Input 1:

-123456789

Sample Output 1:

Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu

Sample Input 2:

100800

Sample Output 2:

yi Shi Wan ling ba Bai

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define Max 300
using namespace std;
struct user
{
	char name[11];
	char pas[11];
}a[Max],b[Max];
char getc(char a[],int m)
{
	return a[strlen(a)-1-m];
}
int main()
{
	int m,m1=0,m2=0,m3=0,k1,k2,k3,cc=0;
	int a[11],b[11],c[11];
	char z[5][10]={"Qian","Bai","Shi"};
	char s[11][11]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
	for(int i=0;i<11;i++)
	{
		a[i]=b[i]=c[i]=0;
	}
	scanf("%d",&m);
	if(m<0){
		printf("Fu ");
		m=-m;
	}
	m1=m/(100000000);
	m2=(m-m1*100000000)/10000;
	m3=m-m1*100000000-m2*10000;
	k1=m1;k2=m2;k3=m3;
	if(m==0) printf("ling");
	else {
		if(m1!=0)
		{
			cc=1;
			int k=0,x=0;
			while (m1>0)
		{
			a[3-k]=m1%10;
			m1=m1/10;
			k++;
		}
			for(int i=0;i<4;i++)
			{
				
				
				if(a[i]!=0)
				{
					
					printf("%s",s[a[i]]);
					
					if(z[i]!=NULL&&i<=2)
						printf(" %s",z[i]);
					if(a[i+1]!=0||a[i+2]!=0||a[i+3]!=0) printf(" ");
				}
			}
			printf(" Yi");
		}
		if(m2!=0)
		{
			
			if (k1!=0) printf(" ");
			int k=0,x=0;
			while (m2>0)
		{
			b[3-k]=m2%10;
			m2=m2/10;
			k++;
		}

			for(int i=0;i<4;i++)
			{
				if(i>=1)
				{
					if(x==0&&cc!=0){
					printf("ling ");
					x=1;
					}
				}
				if(b[i]!=0)
				{
					cc=1;
					x=1;
					printf("%s",s[b[i]]);
				
					if(z[i]!=NULL&&i<=2)
						printf(" %s",z[i]);
						if(b[i+1]!=0||b[i+2]!=0||b[i+3]!=0) printf(" ");
				}
			}
			printf(" Wan");
		}
		if(m3!=0)
		{
		  if(k2!=0||(k2==0)&&k1!=0)	printf(" ");
			int k=0,x=0;
			while (m3>0)
		{
			c[3-k]=m3%10;
			m3=m3/10;
			k++;
		}
			for(int i=0;i<4;i++)
			{
				if(i>=1)
				{
					if(x==0&&cc!=0){
					printf("ling ");
					x=1;
					}
				}
				if(c[i]!=0)
				{
					x=1;
					printf("%s",s[c[i]]);
					
					if(z[i]!=NULL&&i<=2)
						printf(" %s",z[i]);
					if(c[i+1]!=0||c[i+2]!=0||c[i+3]!=0) printf(" ");
				}
			}
		}

	}
	system("pause");
	return 0;
}